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<math> \left | H(e^{jw}) \right \vert (Magnitude) = \frac{1}{N} \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}}</math>
 
<math> \left | H(e^{jw}) \right \vert (Magnitude) = \frac{1}{N} \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}}</math>
  
<math> \angle H(e^{jw}) (Phase) =  e^{-jw(\frac{N-1}{2})}   (when \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} > 0)</math>
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<math> \angle H(e^{jw}) (Phase) =  -w(\frac{N-1}{2})  (when \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} > 0)</math>
  
  
  
*Warning:The answer is completed by one of the ECE 438 Students, not by Professor. The answer is not guaranteed whther or not it is right.
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*Warning:The answer is completed by one of the ECE 438 Students, not by Professor. The answer is not guaranteed whether or not it is right.
  
  
  
 
--[[User:Kim415|Kim415]] 06:03, 4 February 2009 (UTC)
 
--[[User:Kim415|Kim415]] 06:03, 4 February 2009 (UTC)
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[[Category:ECE438Spring2009mboutin]]

Latest revision as of 04:54, 4 February 2009

Homework 2 Question 5.a

$ Y(e^{jw}) = \frac{1}{N}(X(e^{jw})+e^{-jw}X(e^{jw})+.....+X(e^{jw})e^{(N-1)jw}) $

      $  = \frac{1}{N} \sum_{k=0}^{N-1} e^{-jwk} X(e^{jw}) $
      $  = \frac{1}{N} e^{-jw(\frac{N-1}{2})} \frac{e^{jw(\frac{N}{2})} - e^{-jw(\frac{N}{2})}} {e^{j(\frac{w}{2})} -e^{-j(\frac{w}{2})}} X(e^{jw}) $
      $  = \frac{1}{N} e^{-jw(\frac{N-1}{2})} \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} $


$ \left | H(e^{jw}) \right \vert (Magnitude) = \frac{1}{N} \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} $

$ \angle H(e^{jw}) (Phase) = -w(\frac{N-1}{2}) (when \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} > 0) $


  • Warning:The answer is completed by one of the ECE 438 Students, not by Professor. The answer is not guaranteed whether or not it is right.


--Kim415 06:03, 4 February 2009 (UTC)

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